| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Moderate -0.3 This is a straightforward moments problem requiring students to identify that tipping occurs when the person stands beyond support Q, then check if their moment about Q exceeds the beam's restoring moment. The calculation involves only basic moment equilibrium with clearly defined positions and masses, making it slightly easier than average but still requiring proper understanding of the tipping condition. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Taking moments about Q: \(50g\times 0.3+R_p\times 2.1=4g\times 0.9\) | M1 | Finding moment of a force about any point; if weight of person shown between P and Q allow M1 maximum |
| Correct equation from moments about Q | A1 | Could also be obtained from moments about P and resolving to evaluate \(R_p\) |
| \(R_p=\frac{3.6g-15g}{2.1}[=-53.2]<0\), so the beam will tip | A1 [3] | Conclusion must be clear from correct working (\(R_p\) need not be evaluated but must be clear that it is negative) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If person stands \(x\) m beyond Q so beam is on point of tipping \(R_p=0\); taking moments about Q | M1 | Finding moment of a force about any point |
| \(4g\times 0.9=50gx\) giving \(x=0.072\) | A1 | Correct equation from moments about Q |
| \(0.072<0.3\) so this is possible while standing on the beam | A1 | Conclusion must be clear from correct working and reference to 0.3 m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Largest clockwise moment of weight of person about Q is \(50g\times 0.3=15g[=147]\) | M1 | Finding moment of a force about any point; SC 2 (omitting \(g\)): \(0.9\times 4<0.3\times 50\), \(3.6<15\), so beam will tip |
| Anticlockwise moment of weight of beam is \(4g\times 0.9=3.6g[=35.28]\) | A1 | Both correct moments about Q required |
| The moment of the person's weight is larger so the beam will tip | A1 | Conclusion must be clear from a comparison of moments of two forces |
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Taking moments about Q: $50g\times 0.3+R_p\times 2.1=4g\times 0.9$ | M1 | Finding moment of a force about any point; if weight of person shown between P and Q allow M1 maximum |
| Correct equation from moments about Q | A1 | Could also be obtained from moments about P and resolving to evaluate $R_p$ |
| $R_p=\frac{3.6g-15g}{2.1}[=-53.2]<0$, so the beam will tip | A1 [3] | Conclusion must be clear from correct working ($R_p$ need not be evaluated but must be clear that it is negative) |
**Alternative solution:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| If person stands $x$ m beyond Q so beam is on point of tipping $R_p=0$; taking moments about Q | M1 | Finding moment of a force about any point |
| $4g\times 0.9=50gx$ giving $x=0.072$ | A1 | Correct equation from moments about Q |
| $0.072<0.3$ so this is possible while standing on the beam | A1 | Conclusion must be clear from correct working and reference to 0.3 m |
**Second alternative solution:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Largest clockwise moment of weight of person about Q is $50g\times 0.3=15g[=147]$ | M1 | Finding moment of a force about any point; SC 2 (omitting $g$): $0.9\times 4<0.3\times 50$, $3.6<15$, so beam will tip |
| Anticlockwise moment of weight of beam is $4g\times 0.9=3.6g[=35.28]$ | A1 | Both correct moments about Q required |
| The moment of the person's weight is larger so the beam will tip | A1 | Conclusion must be clear from a comparison of moments of two forces |
---4 Fig. 4 shows a uniform beam of mass 4 kg and length 2.4 m resting on two supports P and Q . P is at one end of the beam and Q is 0.3 m from the other end.\\
Determine whether a person of mass 50 kg can tip the beam by standing on it.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{59e924e6-8fa9-4035-9173-705fce487bd9-4_195_977_1676_262}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q4 [3]}}